text:
Where must the glider tow line connection be?
I have styrofoam glider and I want to pull it with a rope, to fly with me when driving a bike. Where do I need to connect the attachment point, if I want the glider to fly stable above me?
I think, to stop roll disturbance I need connect the rope at both wings, but where do I need place them, at the center of gravity (CG), at the center of the overall lift, at 1/4 wing chord (center of wing pressure), or at the aerodynamic center?

(I also notice that a plane tow real glider connects at the nose and when the winch tow connect at fuselage under wing, but what does this place represent in an aerodynamic sense?)

Just connect it at the nose. Also beware phugoid cycle. Will happen if you drive too fast

Why do you think roll will be a problem?

@Abdullah Glider dont have active ailerons to control roll,when I connect to nose only ,glider is roll in circle all the time..

@Abdullah  Why do you think roll will be a problem? What will give my glider inherent roll stability?

Does your glider have roll instability?

I means, youreygonna tow this with a bike so

@Abdullah  yes glider has roll instabillity ,just like every other plane without active controls..

This is simply not true. I believe I have flown your glider. It handles fine. Besides, I can guess from the looks that it is stable and will right itself

@Abdullah What will right him, what give him inherent  roll stability if dont have active ailerons?

How are you trying to start the launch?  Are you holding the glider in one hand and then letting a line reel out after you reach an appropriate speed on the bike?

@Abdullah -- re "This is simply not true. I believe I have flown your glider. It handles fine. Besides, I can guess from the looks that it is stable and will right itself " -- see my answer for some comments on this

It needs to go just under the tip of the nose to have any chance.  This connection is equivalent to the thrust line of an engine, except that because the connection isn't rigid, the longitudinal location matters; the farther forward, the more stable (if the attachment is rigid, like an engine/propeller, it doesn't matter whether the thrusting component is forward or aft, from a stability perspective).
Probably most normal gliders hook up at the nose but a lot of them have hooks near the center of gravity underneath to facilitate winch launching, and when used for aerotow these are less stable and somewhat of a handful to fly for an inexperienced pilot, especially on crosswind takeoffs.
You will have to set the horizontal tail to apply enough trimming force to overpower the nose down component of the pull of the tow line, that is, more nose-up trim than necessary for free flight.
You will find that because the thrust line is more or less concentric with the roll axis, there is not much you can do for roll stability and will be stuck with whatever inherent stability is there from the dihedral effect of the turned up wingtips. Connecting the tow line to each wing tip will not help with roll, since the two lines still run to a single attachment (until you get the angle of the tow line to the longitudinal axis high enough that the "Y" connection is influencing roll, but the thing may still just arc down into the ground anyway), and connecting that far aft will make you completely dependent on the glider's inherent pitch stability and damping.
My prediction is that as is, whether you hook it at the nose or the wing tips, it will go all over the place, than crash, and to get it to just kite along smoothly behind while on tow, you will need a lot of wing dihedral, similar to a 2-control model.

If I hook at c.g.   then I dont need extra force from tail,so glider will fly at lower speeds

@John K Yes you were right,glider is imposible to tow even lines are at tips..Glider roll all over the place and crash...

@JohnK -- a minor quibble/ possible suggestion-- I'd suggest that a CG hook does not in any practical sense offer better handling than a nose hook during aerotow, although yes the responses to yaw and pitch inputs will be somewhat stronger with the CG hook.  My answer gives other reasons why some gliders lack a nose hook.

The tow pilot has to be extra vigilant when launching gliders on CG hooks, and crosswind takeoffs are a challenge.  When towing I've had to cut off gliders in the early part of the takeoff because he drifted off and couldn't recover and was still struggling with it as it was pulling the tug's tail around.  I haven't flown a glider using a CG hook myself tho.

For aerotow, the nose hook has advantages, but for winch-launched free-flying model it would keep the nose down, so the model wouldn't climb. Hook below and just slightly ahead of CG should be the right solution for that (I did see that on a free-flying model).

Downvoting because of the erroneous-- and IMHO dangerous-- claim about CG hooks being used to enhance maneuverability under aerotow. CG hooks are necessary for winch launching, but can be quite the handful when aerotowed. Under aerotow, they have many negative traits: lack of directional stability, lack of pitch stability, and a propensity to back-release when slack develops. I have flown gliders which have both nose and CG hooks, with each hook being used as appropriate for the launching technique. I have flow CG hooks while aerotowed and while it's doable, the downsides are readily felt.

I didn't say it wasn't dangerous. It's obviously dangerous for an inexperienced pilot.  But also obviously, you DO have more maneuverability under aerotow, to the extent that there's nothing stopping you from kiting vertically or coming around nearly alongside the tug, if you were crazy enough or couldn't handle. Some pilots in my club prefer to use CG hook for aerotow because they like the glider not being influenced by the pulling on the rope, as if it had its own engine.  For the tug driver, these gliders had to be monitored much more carefully.

The answer says "have hooks near the center of gravity underneath, for better control authority under tow". This is incorrect. The higher performance gliders are designed for the EU market, which is dominated by winches. Thus, the manuf. prefer to avoid the cost, complexity, drag, and weight penalties of having a nose hook which is seldom used. Esp because, AFAIAA, there is no safe case where we can safely winch launch a glider by its nose hook. So if the manuf. has to choose one hook, it is the C.G. hook because it can serve both safely. Instability under aerotow is not a design target.

Re -- "Some pilots in my club prefer to use CG hook for aerotow because they like the glider not being influenced by the pulling on the rope"  -- this is the first I've ever heard of such a thing.  Disclaimer-- once in a moment of distraction I allowed ground crew to hook the tow rope to the (near) cg hook of a club glider I was flying, that also had a nose hook, for an aerotow flight.  The flight was uneventful, and I didn't realize till the glider was being hooked up for the next flight, what had happened.

@KennSebesta I revised my post.  @ quiet flyer yeah there are a couple of guys who prefer CG hook for aerotow.  They aren't doing it because they think it looks cool.

@JohnK, thanks! I am puzzled by your clubmates, TBH. I'll ask at my club if anyone does the same, but I don't think any of our club or private gliders have dual hooks.

Lots to consider here.  "Free-flight" model airplanes and gliders fly without any control inputs from a pilot whatsoever (and no automatic electronic stabilization systems either).  One branch of this is called "towline gliders" and the "pilot" runs along pulling on a line attached to the glider to get the glider into the air.  The line is attached to a tow hook just forward of the CG of the glider.
If you really want to dive into the subject, you could try to find out just exactly where the tow hook is in relation to the CG for various towline free-flight model gliders.
In full-scale sailplanes, the best location for the tow hook is near the nose for a greater tendency to track straight (and not pitch up) during aerotow, and on the belly near the CG for maximum altitude gain during a winch launch or car tow launch, both of which involve a tow line running to the ground.  Some sailplanes have only one tow hook, rather near the CG, though this is not optimum for tracking straight during aerotow.  Reasons for omitting the nose hook include-- leaving the nose free for something else like a pitot tube, keeping things simpler if there is already a hook near the CG, and keeping things more streamlined (often the hook near the CG is mounted on the landing gear assembly, and retracts with the wheel.)
It is certainly possible to do an car tow launch with a sailplane using just the hook near the nose, but the pilot typically exerts a lot of aft pressure on the control stick, often sometimes getting the elevator into the full up position to allow the glider to really climb steeply despite the downward pull on the nose.  The maximum height gained is always less than on a similar glider with a tow hook near the CG.
It looks like your glider has very modest dihedral (especially in relation to the size of the vertical fin)1 and thus very modest inherent roll stability.  It's doubtful that it could be successfully treated like a purpose-designed free-flight towline glider, in terms of tow hook position.  If you want it to just follow along behind you as you ride a bike, you need to do everything you can to tip the odds in your favor, as far as preventing the glider from turning away and entering a "lockout" (imagine a kite veering "off course", away from pointing straight into the wind, and diving into the ground with string still tight.)  The string should connect rather near the nose-- perhaps 1/3 of the way between the tip of the nose and the CG.  Start there, and then start moving it slowly back to see if you can get the glider to progressively climb a bit higher each time, without going out of control.
Re your idea of connecting the line to both wings-- what about a three-point connection that includes a point maybe about one third or one half of the way out along each wing, and also includes a point on the fuselage closer to the nose?  With the three lines converging several feet below the glider?  It seems that this would "lock" the glider in to a level bank attitude, at least once the glider has gained some altitude, and so long as the glider is staying directly behind the bike.  But it seems that nothing would stop the glider from rotating (yawing) around the towline as it veers off course-- like the "lockout" dynamic we see with a kite on a string, as noted above-- unless the net effect of the three lines is to cause the tow force to converge to a point well ahead of the glider's CG.  It looks like your best bet is going to be to get the tow force to act rather near the nose, in which case the additional lines to the wing are only going to be a hinderance.  You'll likely have better luck trying to just fly it like a glider (i.e. with a single towline), rather than treating it as sort of a hybrid between a glider and a kite.
The Blanik L-13 glider did indeed use a 2-point v-bridle for winch launching and car towing, but the bridle was connected to one point on each side of the fuselage, not on the wings.  See page 7 of this manual for a very brief mention-- note the reference to "very light" control forces, which would have been in reference to pitch control.
Presumably the reason for this design was to get the points of connection absolutely as close to the CG of the glider as possible in the fore-and-aft sense, for maximum altitude gain during car tow and winch launching.  It would not have been possible to put a tow hook on the belly of the glider in the same fore-and-aft location, because it would have been behind the wheel.  Most other sailplanes with a tow hook on the belly of the glider near the CG have it positioned just in front of the wheel.
Please let us know via a comment or an addendum to the question, or a self-answer, what you actually find to work best!
Footnotes--

Some free-flight model aircraft actually have rather modest dihedral, but they invariably have very small vertical fins.  A large vertical fin actually works again inherent roll stability, by minimizing sideslip as the glider is rolling or turning.  The interaction between sideslip and dihedral is key to roll stability dynamics.

if lines connect at same spot(my hand) than glider is imposible to fly..I connect two short lines at each wing(at c.g. line) with steering bar,lines are apart to get roll stability..

@LostInSpace -- ah, that's very different from what I was thinking.  So you have two separated lines.  My answer may not apply very well then.  Is that a new change since you posted the question?  I would think you would probably want them just a little ahead of the CG?  I don't know, maybe at the leading edge of each wing?  Maybe on little rods that project slightly forward of the leading edge?  How long are the lines anyway?

@LostInSpace -- you are going to end up being the expert on this, be sure to tell us when you figure out what works best!

Yes I must put two separeted lines,othervise glider roll over all the time and crash.I mark c.g. line at wings and connect lines directly at that spot...Lines are long enough that glider fly above my head,in clean air.I agree with you, it will be good for pitch stability that lines are little forward from c.g.

Lines(angle between vertical and line,side view) represent direction of glider total aerodynamic force!

@LostInSpace -- well, I would say direction of total force-- i.e. you have to subtract away the weight vector.  Only if the weight of the glider is negligible in relation to the total force on the string, is your idea true. right?  So the faster you pedal, the higher the glider gets, and the closer your statement comes to being true.

yes if weight is negligible


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Where are the small fuel reservoir tanks on a Cessna P210N?
I have looked through the POH and the maintenance manual of the Cessna P210N, both said that the fuel goes from the main tanks into two small reservoir tanks somewhere on the lower side of the fuselage, from there through the fuel tank selector valve to the engine. But neither said exactly where these tanks are located (They could be below the wings, between the engine and firewall, below the pilot's seats, …). Can you tell me where the fuel reservoir tanks are located on a Cessna P210N ?
EDIT: Also, how much fuel fits into each of the reservoir tanks ?

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Do full-feathering constant-speed propellers have a high-pitch stop for the prop governor?
On full-feathering (hydraulic) constant speed propellers in general, and specifically Hartzell and McCauley props, does the propeller governor have a high-pitch stop (like non-feathering propellers, at 20-30°) or can the governor go up to the feather pitch stop if needed ?

Yes the high pitch stops are there, but on a feathering version, they are removed centrifugally when the  RPM gets above 500 after starting. They are only there to prevent the prop from moving into feather when you shut down on the ground.
Actual blade angle is controlled by governor oil pressure driving the blades to fine pitch, opposed by the counterweights on the blade roots wanting to drive the propellers to coarse pitch. On a feathering prop, coarse pitch movement from the counterweights is helped by an internal feathering spring for when it's necessary to go all the way to feather.
The engine feathers by dumping the oil from the governor so that the counterweights and the internal feathering spring are the only forces acting on blade angle, and they move to feather (as long as the engine is spinning above 500 RPM when you feather so that the high pitch stops are disengaged).

What is undesirable about going to feather when shut down or at low speed on the ground? IIRC, the Hamilton Standard props on C-130's (up to the H model, dunno about the J's) could go to feather without limitation on prop speed, but different hardware has different capabilities & limitations. Thanks!

Many turboprops do feather on shutdown.

So the counterweights could drive the blades all the way to the feather pitch if the advance ratio is high enough, right ?

If the oil pressure in the hub is not there, and the prop is spinning, there's nothing stopping the counterweights from driving the blades to feather other than friction and possible opposing aerodynamic forces as the weights move to where their influence is weakest. The feathering spring helps it along to make sure.

@RalphJ Those are "hydramatic" props, where hydraulic pressure drives the blades in both directions. For feathering, they will dump oil from the fine pitch side so oil on the coarse pitch side can drive the blades to feather. Since they are dependent on oil pressure to feather, they use a feathering pump that takes oil from an emergency sump in the propeller gearbox if there is no main oil pressure. TP propellers feather on shutdown to prevent windmilling in the breeze, there being little resistance. Not necessary with a piston engine.

@RalphJ According to my textbook (Power Plants - Oxford): "The feathering springs will gradually push the propeller blades towards the fully feathered position overnight. While this condition would be acceptable on a free turbine turboprop, this would result in an unacceptably high loading on the engine starter motor for a piston engine. To prevent this, centrifugal latches, disengaged with the engine running, will be engaged at an rpm below the manufacturer’s chosen setting, typically 700 rpm."

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What does a DME indicator do in HOLD or RNAV mode?
I found this DME indicator on the Cessna 210. Its mode switch has four positions - NAV1, NAV2, RNAV and HOLD. NAV1 and 2 are clear - it uses the NAV1 and 2 freq respectively. I'm wondering what the other two do, especially HOLD ?
RNAV is also quite logical - it uses the position selected in the RNAV selector. But how does the DME indicator determine the distance/speed to the RNAV position ? By using the GPS ?


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Do aircraft pitch down even at very low angles of attack (below 0°) in a strong sideslip?
I was doing an analysis of an aircraft (specifically the PZL Koliber 160 with the NACA 63A416) strong sideslip conditions (beta=30° for example) in OpenVSP at various angles of attack, and observed that there's a rather strong pitch-down moment even at an angle of attack of -10°. Is that correct, or do I have to go on a long bug-chase in my OpenVSP model ? If it is correct, what causes this ?
P.S.: I did read this post, but my aircraft has a traditional low tail, not a T-tail.
Edit:
I see from the comments and answers I got that my question might not have been clear enough. So - it doesn't matter why the aircraft is flying at alpha=-10° beta=30°, simply assume this as a steady attitude - since that is what OpenVSP does when told to calculate coefficients for that attitude. My only goal is to find out whether the OpenVSP results (specifically the alpha-induced pitch moment coefficient or $Cm_{alpha}$) are valid and thus usable for the FDM of a flight simulator model - that doesn't care either, whether I got into 30° beta by a wind gust, asymmetric thrust, crossed controls or whatever !

An AOA of -15 suggests fully inverted flight. Do you actually mean that or do you mean pitch attitude?

@JohnK no, I mean the angle of attack, and not inverted flight - just a high speed descent with high engine power, for example. And in fact, I meant AoA=-10°, not -15° ! XD

What do you believe the flight trajectory looks like in this condition?

This needs to be tested in a wind tunnel.

Re "no, I mean the angle of attack, and not inverted flight - just a high speed descent with high engine power, for example. And in fact, I meant AoA=-10°"--  if the aircraft is initially upright, then the subsequent flight path will essentially be an earthward (downward) outside loop, with occupants "hanging" in seat belts (heads tending to hit ceiling), and G-meter (if present) showing a negative value. Loop will continue beyond inverted back to upright, and then etc for multiple additional iterations, until strongly negative a-o-a is discontinued. Is this really what you had in mind?

(Not to suggest that the question is inherently defective-- just pointing out that unless we are intentionally doing aerobatics, we would never get into this part of the flight envelope!)

I've been thinking some more about your question and there's a lot to "unpack" here.  If your question is about the pitching moment experienced by the aircraft (with elevator presumably in normal cruise position) then I might suggest editing title to include a reference to "moment".  Otherwise it looks too much like you are simply asking "does the flight path curve down and the nose pitch down when the a-o-a is negative".  Admittedly it is already there in the body of the  question--

@quietflyer and JohnK: I clarified my question - I now see that it was a lot too unclear.

Did you validate your model and have you verified the software gives reasonable output for states that are within the normal envelope?  What is the speed you put in? What pitch moment values does give for that that speed with AoA = 4 and and Beta = 0? Are those values what you are expecting?

@DeltaLima yes, they are. And in fact, I now found a plausible explanation.

Check out the VSPAERO results in my post below -- the chart shows CM vs CL.
CL vs alpha should be a straight line - and CM vs. CL appears to be a straight line (with negative slope).
Together, these facts imply that this result has CMalpha is roughly constant and negative.  Is this what you're seeing?
Negative CMalpha is not a nose down pitching moment.  It is a derivative of a moment.

It seems that this is a phenomenon special to low-wing aircraft, that can be explained by increased drag. In a sideslip, the vertical tail creates a considerable amount of induced drag, which, if the vertical tail's center of pressure is above the center of pressure of the wing - as on a traditional low-wing airplane - creates a pitch-down moment. On a high-wing aircraft this pitch moment is barely present, because the center of pressure of the wing and vertical tail are on roughly the same height.

The relative positions of the centers of pressure does not matter.  What you are thinking about is the position of the center of pressure relative to the center of gravity.
A substantial drag force (apparently on the vertical tail and due to sideslip) above the CG would cause a nose up moment.

As John K was alluding, AoA of -10 or -15 are not really practical and do not correspond to a powered dive like you think.
Even in a descent or a dive, most aircraft will have positive AoA.
You are likely confusing deck angle, angle of attack, and flight path angle.
Angle of attack (alpha) is the angle between the aircraft reference line and the velocity vector.
Flight path angle (gamma or theta) is the angle between the velocity vector and the horizon.
Deck angle is the angle between the aircraft reference line and the horizon.  It is the sum of AoA and FPA.
In a descent or dive, you have a negative flight path angle.  If your descent is steep enough, you may even have a negative deck angle.  You likely still have a positive AoA.
In a steady climb or descent, L = W cos(theta).  CL=L/(q*Sref).  CL=f(AoA).
For negative reasonable angles, cos(theta) is positive and even close to 1.0.  At negative 10deg, cos(theta)=0.98.  In that 10deg descent, your CL is 98% of what it is in cruise!  If your aircraft has symmetrical airfoils and no incidence (i.e. CL=0 at alpha=0), then your descent AoA is 98% of the cruise AoA.  Most importantly, it certainly has the same sign.
It is sometimes counter-intuitive, but climbing, descending, and turning can be treated as 'steady' maneuvers.  For climb and descent, steady means that the accelerations are zero.
Being in a steady climb/descent is separate from the dynamic process of initiating that climb or descent.  That dynamic process is considered maneuvering flight -- unsteady flight.
It is possible to reach negative AoA during a maneuver in a way it is not possible during steady flight.  So an aircraft may experience negative g's during the push-over to initiate the dive -- but in the steady dive, the aircraft experiences flight very similar to level flight.
Edits follow
It would seem unusual for a conventional aircraft to exhibit the characteristics you describe.
Sideslip should have small effect on pitching moment.
A stable aircraft should have a dCM/dCL that is negative -- which implies that at negative alpha it should have large positive CM.
A very quick VSPAERO run on a wing/tail combination shows this.

This shows CM vs. CL for AoA from -10 to +10 and Beta from -10 to +10.
Here is a view of the solution at AoA=-10 and Beta=10.

When building your VSPAERO solution, start simple and build to complex.
A better place to ask OpenVSP questions is on the OpenVSP Google Group.

I'm  not confusing the angles - I do know what alpha is ! ;) I admit however that those -10° are a theoretical value not actually observed on an aircraft, which I picked to cover all corner cases with the results of my OpenVSP analysis. However, this doesn't answer my question - js it normal that an aircraft continues to pitch down very fast even at -10° alpha, in a strong sideslip (> 10°) condition ?

Based on comments as well as original post, the core question appears to be
"is it normal that an aircraft continues to pitch down very fast even at -10° alpha, in a strong sideslip (> 10°) condition ?"
If you are initially flying more or less upright, and then go to an angle-of-attack of minus 10 degrees, yes you will pitch downward very strongly, regardless of what you may or may not be doing with the rudder to create some slip or skid!  You'll note that you are hanging from the seatbelts, with your head tending to hit the top of the cockpit or canopy, as the aircraft begins an outside (downward and earthward) loop.
Note that the net pitching moment determines the change in the pitch rotation rate, not the actual pitch rotation rate.  An outside loop involves some initial nose-down pitching moment, but not a constant nose-down pitching moment, at least if we include the "damping" effect from the curving airflow/ relative wind in our calculation of the overall pitching moment.  Of course, the elevator will have to be deflected in the nose-down direction (in the aircraft's reference frame) throughout the maneuver.
On the other hand, if you are trying to ask if an aircraft will normally experience a strong nose-down pitching moment even with elevator in a normal cruise position if the angle-of-attack is somehow temporarily perturbed to minus 10 degrees, while the aircraft also happens to be sideslipping-- that is an entirely different question.  It would surely be relevant to specify whether the sideslip is caused by deflection of rudder, deflection of ailerons (i.e. via "adverse yaw"), both, or neither (e.g. caused by a sudden transient gust from the side.)


text:
What are typical insurance costs for airliners?
What are typical insurance costs for airliners, and how are they arrived at?
Factors that might be relevant are hours flown, number of landings/takeoffs per year, total hours of airframe and engines, hours on type for the pilot in command, etc. - but are they considered, and how?
Also, are long-haul flights treated differently than short-haul?
Obviously, each specific contract is covered by commercial confidentiality, but some baseline estimates should be available somewhere.

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text:
What is the weight of the batteries on the e-Fan?
Airbus' own (early) page with technical data about the e-Fan provides some information about the batteries.
It can, for instance, be calculated that they hold approx. 19kWh of energy if you accept a somewhat simplistic calculation by multiplying number of cells (120) with capacity (40A ) and voltage (4V).
What I can't find is information about the weight of these batteries. Is this information available somewhere or can it be reasonably estimated?

The weight of the battery is mentioned here:

Power is provided by 127 kg. of 250-volt lithium-ion polymer batteries
  from Kokam in South Korea built into the wing.

A slightly different weight is specified here:

... E-Fan has two lithium-polymer battery packs weighing 65 kg each.

I could not find which Kokam battery is being used on the e-fan, but this page has specifications for all of their batteries.

Slightly adjusting your energy estimate (3.6V × 40A × 120 Cells = 17280Wh vs original 19000 Wh), taking an energy density of 125Wh/kg for a high quality battery, it can be estimated:
$$\frac{17280}{125}=138\text{kg}$$
which I think sounds reasonable for a plane with a empty weight of around 500kg.

Matching this with Farhan's answer, I'd say that's a pretty close estimate!

Nice estimate and thanks for the link, very interesting.

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text:
How do you calculate the deviation from an RNAV waypoint radial?
As a clarification at the beginning: I'm talking about non-GPS, pure VOR-DME based RNAV devices here.
On RNAV devices that use just the selected VOR-DME, not GPS, the phantom radio station is set up by a desired bearing from the VOR-DME station and a distance. Then you can select a desired radial from that phantom station (instead of the VOR-DME station), and the deviation from that radial is then indicated by the CDI.
Given

the aircraft position relative to the station as radial $R_a$ in degrees and distance $D_a$ in nautical miles
the desired phantom station position relative to the VOR-DME station as radial $R_s$ in degrees and distance $D_s$ in nautical miles
the selected radial of that phantom station $R_{cdi}$ in degrees

how would you calculate the deviation $D_{cdi}$ from the selected radial $R_{cdi}$ of the phantom station ?
EDIT:
This is what I have tried so far:

$x_a = D_a * cos(rad(R_a))$, $x_a$ being the X position of the aircraft relative to the VOR-DME
$y_a = D_a * sin(rad(R_a))$, $y_a$ being the Y position of the aircraft relative to the VOR-DME
$x_s = D_s * cos(rad(R_s))$, $x_s$ being the X position of the selected RNAV phantom station relative to the VOR-DME
$y_s = D_s * sin(rad(R_s))$, $y_s$ being the Y position of the selected RNAV phantom station relative to the VOR-DME
$D_{as} = \sqrt {(x_a - x_s) ^ 2 + (y_a - y_s) ^ 2}$, $D_{as}$ being the distance between the selected phantom station and the aircraft in nautical miles
$B = deg(atan2(y_a - y_s, x_a - x_s))$, $B$ being the bearing of the selected phantom station to the aircraft in degrees
$D_r = sin(rad(R_s - B)) * D_{as}$, $D_r$ being the deviation of the aircraft from the selected radial of the selected phantom station

I've tried using these formulas, but I ended up with nonsense values for $D_{as}$ - 0 if $D_s$ was 90, close to zero for smaller values and  90 for $D_s = 0$. I must say that I was never good in trigonometry … so please, can you tell me what is wrong with theses formulas ?

It’s just triangles and trig. What have you tried. How far have you got?

@Jim I edited my question.

Maybe you could get an answer on Math SE?  Seems like a better fit…

@MichaelHall maybe you are right … but anyways it looks like I just made a mistake somewhere as it's giving plausible values now.

To keep it simple and hide the trig here we can do it with vectors.
Using your subscript nomenclature:
Define Va as the vector from the VOR to the aircraft. (The VOR/DME gives this directly in polar coordinates (Radial, distance).
Define Vs as the vector from the VOR to the phantom station.  This is a given when setting up the phantom station.
Define Vsa as the vector from the phantom station to the aircraft.
Then:
Va = Vs + Vsa
And
Vsa = Va - Vs
The CDI shows angular deviation from the selected radial.
Define a unit vector vcdi for the selected CDI radial.
The dot product can be used to get the angle:
vcdi ⋅ Vsa = ‖vcdi‖ ‖Vsa‖ cos α
Where α is now the angular deviation of the aircraft to the selected CDI radial.
But ‖vcdi‖ ( a unit vector) is 1.
So vcdi ⋅ Vsa = ‖Vsa‖ cos α

Could you please add an example calculation, say, for $V_a = (135°, ~14.14 nm)$ and $V_s = (90°, 10 nm)$ (so $V_{sa} = (45°, ~4.14 nm)$) and $V_{cdi} = (180°, 1)$ - because I don't get how to calculate the CDI needle deflection from $V_{sa}$ and $V_{cdi}$.


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Why don't small aircraft produce tyre smoke when landing, but big aircraft do?
On almost any recording of big aircraft landing, a huge cloud of smoke can be seen. This even applies to the relatively small Cessna Citation. But I never saw such smoke, at least not visible one, on a landing of a, say, Cessna 210.
If it is simply the touchdown speed (140 on a Boeing 777, 60 - 70 for a Cessna 210), what is the decision speed between smoke / no smoke ?

It is not only the speed, but the combination of several effects:

Landing speed: The much higher wing loading of airliners, even at the end of their trip, results in touchdown speeds in the area of 120 - 150 knots while GA airplanes will have only half that speed at touchdown. The kinetic energy of the rotating tire is proportional to rate of rotation squared, so an otherwise identical tire needs to change its energy by a factor of four when you compare airliner to GA conditions.
Tire size: Your average GA tire has much less inertia (the quantity which determines how much energy is needed to spin up the tire) than the much larger and heavier airliner tire. Still, spinup happens by friction between rubber and concrete, with much the same limits on material strength and melting point. Inertia is proportional to mass and radius squared, and if you compare the dimensions of GA and airliner tires, radius grows from 24 cm on a Beech Bonanza (main tire mass is 5.8 kg each) to 68 cm on a Boeing 777 (main tire mass is 125 kg each).
Area load is much higher in airline tires. Just compare the typical tire pressure to get an idea how much more weight per area the airliner tires have to carry.
Landing technique: Whereas you want a firm touchdown with airliners so the load on the landing gear will allow spoiler deflection and brakes can immediately be effective (resulting in a spinup time of just 0.07 seconds for airliner tires), the much shorter landing distances of GA airplanes allow to strive for a "greaser" at touchdown, where wheel spinup is gradual and the heat at the contact area of the rubber surface will stay well below its melting point.

As a result, the wear volume (the volume of rubber used up) in a GA landing is insignificant while it reduces the number of landings between tire changes for airliners to just 200. This study shows in detail how wear volume can be calculated while this study goes into the dynamics of how landing smoke is caused for different pre-rotation speeds.

Also consider that most light aircraft pilots have to pay for their own tires :-)  Though you can get some smoke when practicing short field landings.

I'm surprised by your last bullet point. I always thought a firm touchdown spins up the tires faster and therefore produces less smoke (and vice versa).

and of course the smaller size would mean a smaller amount of smoke if there is smoke and it being less visible to an observer (and visible for less time, especially as it's also blown away more by the prop).

@Bianfable potentially, less smoke but all in the same place would be more visible, than a very slow stream of smoke

Have they considered, on large aircraft, adding motors to get the wheels spinning at approximately the correct speed prior to landing?  The motors would add some expense, but might save a lot in the long run in constant replacement of tires...

@DarrelHoffman See this question: Why are aircraft tires not pre-spun prior to landing to preserve them?

@DarrelHoffman I don't have any numbers, but I would be really surprised if tyre wear during the mile-long rollout was negligible relative to the initial spin-up wear (thus eliminating the latter wouldn't really help that much). Consider also the infernal environment into which you would be putting those motors, right next to the brakes dissipating megawatts of heat.

@TooTea The worst tire wear occurs in a rejected takeoff. All other cases are negligible in comparison, so tires are sized for this case. This would also stress those motors the most.

The smoke produced from a stationary tire of, say, 12 inches diameter, contacting pavement at 70 kts, is not enough to be clearly visible because there isn't enough heat energy created during the contact and spin-up.
You have two big factors at work; the contact velocity is low, and the wheel's inertia is low, so that the tire is able to accelerate to its rolling rpm quickly and doesn't spend much time skidding along.
The 777 tire on the other hand contacts at 140kt or whatever it is, and spends maybe half a second to a second spinning up to the full rolling velocity.  That's a lot of sliding rubber at very high velocity, lasting long enough to make a lot of heat.
If you are landing a light aircraft and touch down fast and get hard on the brakes, you can get the tires to smoke quite nicely in some cases.


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Does throttle equal power on a turbocharged piston engine?
On a normal piston engine powered airplane, throttle equals power. If I set the throttle to 50 %, I will get 50 % of the power the engine can develop at the current altitude, temperature, etc. In a flightgsimulator, I am getting full power and manifold pressure from 75 to 100 % throttle at sea level under ISA conditions. Is that normal ? I was expecting to get the same results as with a piston engine, i.e. that 50 % throttle produces 50 % power, and 100 % throttle produces 100 % power.

What do you mean in your last sentence?  Are you asking whether you will get full power if you are at 65% throttle?  (i.e. power output flat at all throttle settings from 65 - 100%?)  Because on the surface this makes no sense...

@MichaelHall I edited my question - I hope it's clearer now.

That's a bit better, but what are you basing your throttle % on?  (lever position, RPM, manifold pressure?)  And what are you using to gauge power output?  The short answer is that there is no compelling reason why a turbo power curve would go flat at 65% of anything, so this is probably a coding bug.  But I'm sure someone here will post up some HP and torque curves for turbo vs normally aspirated engines for a more complete answer...

@MichaelHall the throttle position is set directly as a fraction (so 0.65 for 65 % throttle(, and the output power can be read as horse power, RPM and manifold pressure.

@MichaelHall no I don't think it's off-topic, because I want to know how this behaves on the real thing.

@MichaelHall we have a flight-simulator tag.

Is it turbocharged or turbonormalized?

@Jim turbocharged with a maximum manifold pressure of 36.5

As John K correctly explains, the relation between throttle position, or even manifold pressure, and torque or power is never linear. But I'd like to add one thing:
If the engine power, and the manifold pressure, stops increasing at 65% (or whatever other %) of throttle movement, you have an engine with manifold pressure red line, protected with automatic waste-gate (a “flat rated” engine, though that term only became common later with turbines).
The point of turbocharging an engine is that it can produce more power at high altitudes. So say your engine has turbocharged capable of producing 45 inHg manifold pressure up at 19,000 ft. The manifold pressure is still proportional to the density altitude, so at sea level, if you opened the throttle wide, the manifold pressure would rise to around 81 inHg. But if the engine was to handle that, it would have to be too heavy, and spark-ignition engines (unlike diesels) are limited to around maybe 55 inHg anyway because above that the air charge will get too hot and trigger preignition, which reduces the power and damages the engine.
So with such engine, you are told not to let it run over, say, 54 inHg manifold pressure, or the engine might be damaged.
The newer engines have automatic waste-gates in the turbo. Those are valves that will relief the pressure if it exceeds the red line. So if the engine hits that 54 inHg MP at 65% throttle, and you keep advancing the throttle, the waste gates will open and the manifold pressure will remain around 54 inHg, and the engine will not produce any more power.
Older engines did not have waste-gates, or only manually operated ones, so you had to be careful not to overboost (exceed the MP) the engine.
With big engines you are also likely to be told not to exceed some lower limit for more than some time. For example max 2 minutes above 49 inHg. That the pilots always have to watch for themselves.

Yep … the engine in question actually is described as flat rated. This was the answer I was searching for ! Thanks and +1 ! But one question - if the throttle controls the fuel flow (that's what I thought - correct me if I'm wrong), what happens to the excess fuel when the wastegate opens ?

@Programmer, throttle, on a spark-ignited engine controls the throttle, a valve restricting the intake manifold. Fuel is injected later, metered according to the air flow into the cylinders to create (approximately, adjustable with the mixture lever) stoichiometric mixture.

Throttle position in itself doesn't equal power on any piston engine.  Power is a torque vs RPM, as in any engine.
Aircraft piston engines, whether turbocharged or not, use manifold pressure, the absolute pressure upstream of the throttle plate, given in inches of mercury like a barometer, as a proxy for torque, and the pilot will use a chart with curves that plot RPM and manifold pressure to derive horsepower being produced.  In cars, if manifold pressure is being monitored, they reverse the concept and express it as vacuum (the differential between ambient and absolute pressure in the manifold), since the pressure in the manifold is normally lower than ambient, absent supercharging/turbocharging.
For a given atmospheric condition, manifold pressure will vary with throttle opening. Because power is a function of torque and RPM, 50% power will be achieved within a band of manifold pressures (throttle settings) and RPM values; for example, you may get 50% power at 20" of manifold pressure and 2000 RPM, or 19" of manifold pressure and 2100 RPM, or 21" of manifold pressure and 1900 RPM.
If the propeller is fixed pitch, you lose the ability to control RPM, and there will be one specific throttle setting that results in 50% power, the point where the manifold pressure and RPM add up to 50% of rated power for a given ambient condition. This setting however will vary with the atmospheric conditions and altitude, so you still don't end up with a throttle position that represents 50% in all conditions.
Turbocharging/supercharging doesn't really change things other than the fact that manifold pressures may be higher than ambient at maximum power, and certain manifold pressures can be achieved at high altitudes that a normally aspirated engine can't.  As well, on a turbocharged or supercharged engine, you will have a maximum throttle setting that is a function of a manifold pressure redline, which can be less than physically wide open throttle.
On a PW-985 radial, which has a mechanical supercharger, the maximum throttle setting is at 36" of manifold pressure. Pushing the throttle all the way forward might get you 38 or 39", which will damage the engine.  So in that case, you advance  throttle to the MP redline and stop there, and you will be somewhere short of the throttle's physical travel limit, whereas on a normally aspirated engine you just push it until it stops.
It still comes down to the fact that X% percent power is achieved at various combinations of manifold pressure and RPM, and the throttle opening that results in a given manifold pressure will vary accordingly and there isn't a single universal setting except for wide open throttle.

Hmm. But this doesn't really answer my question if it is normal that I get 100% power from 70% to 100% throttle.

You will get 100% of rated power IF throttle is wide open, redline manifold pressure is achieved at WOT, AND propeller is regulating RPM at redline.  Throttle must be wide open to get 100%, but only if the other conditions are also met.  If they are not, you will be at something less than 100%.

Yes, but does 70% throttle already count as "wide open" ?

Oh I see what you mean.   Edits made.  Yes on a turbocharged engine max power is usually less than wide open throttle, and you set power based on a MP redline.

Flight simulators used for aircrew training must replicate essential flight control behaviour, of which the throttle response is an important one. In flight sims this is implemented as follows:

Measure the input parameters and associated output parameters (fuel flow), in multiple relevant circumstances. Such as in this Swiss research project on aeroplane emissions of a carbureted engine, in which they do state that "Piston engine fuel flow is constant for a given amount of power".
Design system functionality based on the physics of the system. For the fuel system that includes fuel temperature & compressibility, pump behaviour, friction in the supply lines etc.
Program the system functionality, then calibrate using the data points from the measurements.

A statement like “On a normal piston engine powered airplane, throttle equals power” would definitely need to be verified with measurements. Is the power setting really linear with fuel flow?
However, the behaviour you describe could be the effect of the characteristics shown in this answer: turbocharged and supercharged engines do have a non-linear relationship between max power and altitude.

"When the throttle is at 0%, does the engine switch off?" - no, of course not. But if I get for example 10% power at idle throttle (0%), I get 55% power at 50% throttle and 100% power at 100% throttle (theoretically). The question is if this is also the case for turbocharged engines, or if it is normal that I do not get any power change between 70% and 100% throttle - i.e. the engine running at maximum power at any throttle setting between these two. Or said in another way - with a supercharged engine, does the throttle control turbocharger RPM or fuel flow / engine RPM ?

Depends on the aircraft! Have you got any manufacturer data?

The specific aircraft I want to know this of is the Cessna Pressurized Turbo Centurion P210N, equipped with a Continental TSIO-520-P engine. I have only the POH of the aircraft - for the engine itself I could only find some small bits of data - stroke, bore, etc., nothing about such details.

The flat power curve at loower altitude of turbocharged pistons can indeed cause that effect, yeah.


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What is the maximum control surfaces deflection on a Cessna 210N?
Does someone know how far the control surfaces (rudder, ailerons, elevator and nose gear) can deflect in a Cessna 210N? I've been searching Google and the POH for hours now and can't find it.


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What is the round tube mounted on the roof just behind the windshield of a PC-6?
If you look at the below image:

(image source: Wikimedia Commons)
There are three "antennas" mounted on the forward portion of the cabin roof. The two rear, wing-shaped ones probably are the COM1/2 antennas, but I'm clueless as to what the the straight vertical tube (circled in red) mounted directly behind the windshield could be. Can anyone tell me what it is ?

Close-up pictures of the object in question:
https://i0.wp.com/modeltalking.hobbyart.sk/wp-content/uploads/2015/08/IMG_0644_vyrez.jpg
In all pictures I could find, it is painted in the same color as the fuselage, which leads me to believe it is not in fact an antenna.

thanks @Wasserwaage for the picture and the bounty ! :) I don't think it's an antenna either, but I didn't know  how to better describe it.

The pictures of older PC-6 models here: http://www.pc-6.com/portergallery.htm show this object mounted on planes manufactured way before 2003, which was apparently the first year that ADS-B was used in general aviation: https://en.wikipedia.org/wiki/Automatic_Dependent_Surveillance%E2%80%93Broadcast#United_States

The vertical tube on top of the cabin is the fuel system vent mast:

Pictures are taken from an online copy of Royal Australian Air Force Pilatus Porter Flight Manual, Australian Air Publication 7211.001-1

Almost posted a guess to that effect!  Then figured "no, an airplane with wing tanks probably won't be vented at the centerline"-- wrong!

I was sure it had to be either some weird static port mast, or a fuel vent. Oddly enough it's not easy to find PC-6 POHs online despite it being such an long lived and common plane.

ooooh, the fuel vent … thank you very much ! :) How do I award the bounty ? There's no bounty award button as mentioned in the help article ! :(

@quietflyer oh okay, thanks !

@Programmer You don't get to award the bounty. Since qq jkztd started the bounty, they will be able to award it. The button is under the checkmark (only visible to qq jkztd at the moment).

@Bianfable oh okay, makes sense that I can't award someone else's bounty - thanks for explaining ! :D


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Why does stalling increase drag?
I understand (I think) why lift reduces in a stall.
What I don't understand is why drag increases. I can see that the ratio of drag to lift would increase,  but I'm not sure I understand the mechanism that leads to an absolute rise in drag.
Lots of sources mention it, but they just refer to turbulence or vortices or similar without really explaining how those phenomena cause the drag to increase.
Is this an increase in induced drag or parasitic drag? I think I could understand if it was just increased parasitic drag.

Induced drag is drag that results from the production of lift. Since lift drops sharply in a stall, induced drag would also drop. Parasitic drag does increase. Is your question about gaining an intuitive "macro scale" understanding of why high angle of attack generates parasitic drag? Or more about the nitty gritty physical aerodynamic processes that result in a net force?

Uh, both I think. But more the second one than the first.

Stalling means that flow separation increases so lift does no longer grow with angle of attack. This flow separation happens mostly over the rear, inner part of the wing, where the local surface inclination has a backward component.
Now compare the pressure over an airfoil for fully attached and partially separated flow, just as it looks near or at stall. The plot below shows an XFOIL result for an airfoil at 12° angle of attack. The dashed lines show the inviscid pressure distribution (the one without separation) while the solid lines show the viscous pressure distribution with separated flow over the last 20% of the upper side. If you need more information to interpret the plot, please consult this and this answer.

Note that the scale on the y axis is negative: A more negative value means more suction. The stalling airfoil shows lower suction over most of the chord and only over the last 10% its local pressure is lower than it would be without separation.
This means that stall reduces suction over the area of the wing which points forward (there is really the expression nose or leading edge thrust for this suction) while it increases suction over the part that points most backward. In sum, both effects increase drag.

Is this an increase in induced drag or parasitic drag?

Actually, it is a lift deficit due to flow separation while induced drag still grows. Near stall more lift has to be bought by a disproportionate increase in angle of attack which brings with it an equally disproportionate increase in both induced and parasitic drag. Post-stall, lift drops again but drag continues to rise.

Drag is just the component of the aerodynamic force on the entire airframe that is aligned with the aircraft's velocity Vector. (as lift is the component normal to or perpendicular to the velocity). When an aircraft stalls, the vectoral summation of all aerodynamic forces is pointing aft, with a much greater angle (think about pushing a door through the air or holding your hand out the window of a car).
Drag can be broken up into multiple individual components (like form drag, induced drag, etc. etc.),  but the simplest answer is just that when the wing stalls, the total force is pointing backwards at a much greater angle than it is when the aircraft is not stalled, and cutting through the air cleanly.

Stalling increases drag dramatically indeed. The pic above is the $C_D$ over 180° from this answer, which also contains the corresponding graph for the $C_L$. At about 12° we can see the huge jump in $C_D$, after which the curve follows the shape for flat plate lift.
The jump in drag is particularly drastic for the NACA 0012 because it falls in the category of profiles with leading edge stall, as described in this answer: a thin bubble develops over the leading edge and then pops open at the critical stall angle. Thick airfoils have trailing edge stall with a more gradual behaviour.
The sudden increase in drag of a fully statically stalled profile is parasitic in nature. The separated flow from the upper surface produces a reduction in lift (induced drag), and much more drag than the friction drag from a nicely attached flow.


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How would you adjust C_l airfoil values for different Reynolds numbers?
I have graph of C_l (airfoil lift coefficient) as a function of AoA, for Mach 0.7 and Re = 6x10^6.
For a given value of C_l, what would be a good way to estimate that value at a different Reynolds number. 6 million is quite low, most transport jets fly at significantly higher Reynolds numbers.

Normally, the Reynolds number affects lift very little, except near maximum lift. Could it be you want to ask about drag instead?

@PeterKämpf I'm looking to adjust C_lmax values precisely.

The following two pictures taken from the classical NACA's “Summary of airfoil data” by Abbott, Von Doenhoff and Stivers depict the trend of $C_l$ and $C_d$ in respect of $Re$:

Reynolds number is defined as the ratio of the inertial force possesses by the airflow in relation to the viscous (shear) forces within it:
$Re=\frac{\rho Vc}{\mu}$
This can be better seen multiplying both numerator and denominator by $V$ and rewriting the definition of $Re$ as:
$Re=\frac{\rho V^2}{\mu(V/c)}$
Now the numerator resembles the classical expression for kinetic energy while the denominator represent the shear force given by the viscosity $\mu$ due to a speed variation $V/c$.
So, big Reynolds numbers imply bigger inertial forces than viscous ones. Small Reynolds numbers imply bigger viscous forces than inertial ones.
Intuitively that means that at small Reynolds numbers, viscous characteristics are predominant: thicker boundary layer, bigger viscous drag and earlier stall i.e. lower $C_{l_{max}}$.


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What are engineers trying to maximize when they design for takeoff at MTOW? (It seems to me the only free variable is the OEW.)
For a commercial flight, the payload it carries is the weight of passengers and their baggage. Therefore fixed by seat configuration. Fuel weight is determined by the range of the flight. MTOW or maximum structural take-off weight.
So what exactly are engineers trying to maximize when they design for take-off at MTOW?
MTOW = OEW + PL + FW. - This expression seems to indicate that the only free variable is the operating empty weight, and that's what they are trying to maximize, but obviously that doesn't make sense.

My question is why most planes (and most airlines do it this way) are designed to take off on the MTOW line in this PL-R diagram (2-3), and not on the maximum payload line (1-2) or on the maximum fuel line (3-4).

Thanks for the reply. Unfortunately that question seems to address a different issue.

MTOW as defined in the airport planning manuals, which refers to the structural limit.

Hi Bianfable. I mean that that is the formula for the MTOW line on the PL-R diagram. My question is why are airplanes designed to take off at this weight and not at MPL or MFW. Most airlines take off at MTOW and never at MPL or MFW.

@MikeSowsun It's possible to take off at MPL or MFW (see PL-R diagram I uploaded), planes do it under special circumstances. At MPL you reduce the fuel to stay below MTOW, at MFW you reduce payload. My question is more about the motivations of airlines to choose the middle line of this diagram. MLW could be an answer for short routes I suppose, but it wouldn't answer the question for long routes.

Why MTOW is particular I guess is what I'm asking. Why don't planes take off underneath that maximum.

Commercial airliners routinely take off well below the Max weight. They even reduce the thrust on the engines below max thrust for most takeoffs. Why do you think otherwise?


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What are typical values of CD0 for commercial high-subsonic jets?
Torenbeek (engineer famous for his book on aircraft design) indicates 0.014-0.020. However I've been estimating values for typical planes like the B737/A319 and I'm getting values of 0.026-0.027. I'm interested in planes of approximately this size. This makes me think that Torenbeek may be outdated.
Does anyone know typical values? CD0 being the parasitic drag coefficient of the entire aircraft.

Torenbeek is right, the high Reynolds numbers of large and fast transports make such low values possible. It would be interesting to see the basis of your results; without those I cannot really give a better answer.

The $C_{D0}$ table in Torenbeek's book. The version I have was published in 1982, at the time the B737-200 had been in operation for 11 years. Plus the following high-subsonic jet aircraft:

Airbus 300 since 1974
B-707 since 1958, including the stretched B707-320 operational since 1959
B-727 since 1964, including the 727-200 operational since 1967
DC-8 since 1959, including stretched versions through the DC-8-63
DC-9 since 1965, including stretched versions through the DC-9-50
F28 since 1967, including stretched versions through to 1973.
Many others, such as the B747, Ilyushin 62, L-1011, Caravelle, Comet, BAC 111 etc.

The $C_{D0}$ values in the table were based on all high subsonic jets in operation at the time. Note that the B737 shares the fuselage diameter with the B727 and B707 - length of the current B737Max at 43.8m is still a bit shorter than the B707 @ 46.6m, so is the wing span. The 4 narrow cross section engines of the 707, versus the dual engines  with much wider cross sections of the Max.
Nothing in the configuration changes since 1982 strikes me as a probable cause of an increase in $C_{D0}$ to that of a twin engine piston aircraft.

@ymb1 Thx for the doc, interesting info. They use statistical methods to estimate a matrix of parameters, the results have quite an uncertainty spread, not in the least because they don’t know the actual mass. At the time of writing of Torenbeek the manufacturer windtunnel data was much more freely available.


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Is there a relationship between thr increasing number of airliners flying over 11000 metres and the number of problematic air pocket events?
I saw many news about the death of a passenger on Singapore Airlines flight. They say that the accident was caused by a turbulence. But judging from the description of what happened I thought about an air pocket. I was surprised when I saw that the plane was flying so high, I still associate the airliner with a plane that flies at a altitude slightly above 10000 metres.
I had a look on Flight Radar, I chose some random long range flights and I saw that most of them were travelling above 11000 metres and some above 12000 metres. It seems that the average travelling altitude is increasing. Is it going to have an impact on the number of events of planes dropping by a thousand metres or more due to air pockets during the flight?

https://skybrary.aero/articles/clear-air-turbulence-cat

"Air pocket" and turbulence mean the same thing.

Airliner cruise altitude is not increasing, at least not in the past forty years.

Please cite your source that “… the average travelling height is increasing.”

FWIW, the cause of death is suspected to be a heart attack:  https://en.wikipedia.org/wiki/Singapore_Airlines_Flight_321  It could have been influenced by stress as a result of turbulence, but turbulence was not the direct cause of death.

Boeing models, for example, have had service ceilings in the low 40kft range for decades now.  That said, obviously, over the years the number of revenue passenger miles above (for example) 35kft has increased considerably. To me an interesting statistic would be number of injuries (or deaths) attributed to turbulence per revenue passenger mile, by year.

In short
Aircraft cruise altitude is linked to the tropopause altitude, which rises with climate change, about 1 000 ft in 50 years.
Turbulence events are not linked to this change in altitude but are still linked to climate change. Their rate may double or triple in the coming years, and it is a certainty their intensity will increase.
This is the general trend for many atmospheric events. High inertia oceans warm, leading to more evaporation and more water in air, thus more energy is stored everywhere, thus stronger winds are generated.
Some more details below if you're interested.

Troposphere and tropopause
Atmosphere temperature starts at about 15°C at sea level and continuously decreases at the rate of 1°C per 180 m while moving away from the warmer Earth, until reaching -50 to -65°C at an altitude of about 8 to 17 km (depending on latitude and month).
This layer of the atmosphere where temperature decreases is called the troposphere and its upper limit the tropopause. For a latitude of 45° the tropopause altitude is about 10-11 km
Tropopause rise

It seems that the average traveling altitude is increasing.

This increasing cruise altitude is likely linked to the increasing altitude of the tropopause. Aircraft and their engines are optimized for an altitude  close to the tropopause where the most favorable combination of temperature, density and pressure is found.
The mean atmosphere temperature started increasing circa 1920:

Source. Data source: WMO/AFP. Full size image.
This moderate change inflated a little bit the troposphere, inducing a tropopause rise by 50-60 m (160 ft) per decade. Over the last 50 years the tropopause altitude increased by about 250-300 m (820-980 ft).
Flight levels
Cruising aircraft fly at "levels", which are hundreds of feet, e.g. FL360 is 360 x 100 ft = 36 000 ft.
Aircraft that were optimized for flight level FL340 (10 360 m) in the 70s must now fly at FL350 (10 670 m) to get the same conditions.
Similarly cruising at FL350 (10 670 m) has evolve to cruising at FL 360 (11 970 m), etc, there is a general rise of 1 000 ft or +10 in FL unit.
Clear air turbulence
Turbulence is another word for wind. Given the aircraft own motion, the most problematic impact is from a vertical wind, they should be avoided. Many turbulence areas, often located between the ground and cumulonimbus clouds, can be remotely detected by airborne weather radars and by-passed.
Clear air turbulence (CAT) is another type of turbulence, linked to jet streams, occurring mostly close to the tropopause. It cannot be detected by radars and is the most frequent cause of a sudden, sometimes spectacular, vertical aircraft motion (an obvious reason to have the seatbelt fastened in cruise).
CAT events dynamics

Is it going to have an impact on the number of events of planes dropping by a thousand metres or more due to air pockets during the flight?

The number of CAT has increased, and CAT events are more intense. The intensity is expected to continue to increase, and it can pose an actual threat to commercial aviation in the long term.
According to this study, about CAT over North Atlantic:

Severe-or-greater CAT increased the most, becoming 55% more frequent in 2020 than 1979.

This is a bit less over the US (41%). The trend is not good either:

Turbulence strong enough to pose an injury risk could double or triple in frequency.

Number of injuries (or deaths) attributed to turbulence per revenue passenger mile, by year? Anybody? RPM data is not hard to come by but need help on the other.

Was there some reference to 1920 and trend in that link you provided?

@ThomasPerry: This is a link to the diagram. The baseline for the temperature anomaly is the mean of years 1881-1900, the pre-industrial era where the global temperature (sea and land) was relatively constant. The anomaly figures are the difference with the baseline. You can see on the first diagram, the increase indeed starts in 1920, and you can read more on the pre-industrial reference in What is a pre-industrial climate.

Nonetheless, interesting...


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How are sensors in the wheels connected to power and data buses?
In this question the sensors to measure the tyre pressure are described.
But how do such sensors receive the electrical power they need, and how are the signal values transmitted back to the relevant computer systems?
Since the wheel rotates, one would presume that transferring power and data electrically with the reliability needed for aviation use would pose specific demands on the technology.

For small planes simple systems can be installed that fit on the inflation tubes and replace your standard cap. They use small batteries in the sender units and a little read out.
It looks like Boeing gets their tire pressure/break temp systems from Crane Aerospace. The latest and greatest systems are all wireless

The Crane 777 Tire and Brake Monitoring System (TBMS) provides
  continuous real-time monitoring of aircraft tire pressure and brake
  temperature. The design incorporates our SmartStem® wireless tire
  pressure sensor with integrated fill valve. The SmartStem uses sensing
  technology with triple redundant sensing channels to provide high
  reliability and accuracy. The Crane TBMS replaces TPMU and BTMU units
  (from the legacy system) with a single Tire and Brake Monitoring Unit
  (TBMU) that interfaces with the existing brake temperature probes. The
  existing wheel interface unit is replaced with an in-axle transformer
  and relay transformer hubcap assembly that allow the tire pressure
  signal to be transmitted magnetically with no physical contact between
  the components. Retrofit of the Crane TBMS requires no change to the
  AMC, EICAS, and CMC system interfaces.

That does not shed to much light on how it all works though. Their handheld system is RFID based and presumably their built in system is similar.


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Why were the speed records set by seaplanes in the early thirties?
For nearly all of the 1930'ies, the fastest aeroplanes were all of the seaplane category (as opposed to landplanes, to use the terminology of FAI).
Presumably, the designers of the two categories had access to more or less the same technology (engines, materials, aerodynamics, etc.) given the large interest in the subject and funding available.
If we for a second rely on Wikipedia for the record listing, we see that the speed record for a seaplane set in 1931 is 655.8 km/h (Supermarine S.6B), while the record for a landplane set in 1935 (4 years later) is "only" 567.1 km/h (Hughes H-1 Racer).
Seeing how fast aviation technology developed in that period, it is remarkable that four years after the Supermarine record, the fastest landplane was still significantly slower than the fastest seaplane.
I would expect the large pontoons on the seaplanes to provide more drag compared to a simple landing gear, and hence by logic the seaplane should be penalised and achieve lower maximum speeds. All things being equal.
However, since the speed differences are so pronounced, all things obviously are not equal. So what are the differences between the two categories (or their designers and manufacturers) that can explain the gap in maximum speed?

Supermarine S.6B's engine: Rolls-Royce R, 2,350 hp (1,753 kW), H-1 Racer engine: Pratt & Whitney R-1535 radial engine, 700 hp (521 kW). So I guess this question boils down to "Why seaplanes manufacturers chose to stick more powerful engines to their planes"

In an era of fixed pitch propellers and limited flap technology a seaplane may have had certain operational advantages compared to the airfield needed to support a landplane of comparable speed-optimized fixed configuration in the takeoff/landing regimes where that configuration is not optimal.

@DeepSpace Interesting observations. The question though would rather be "why did landplane manufacturers not use the same powerful engines that seaplane manufacturers did?"

@ChrisStratton Thanks for the observation. Peter Kämpf's answer goes in the same direction, so this is worth delving deeper into, I guess.

The answer: Fixed-pitch propellers.
Until the 1930s propellers were fixed pitch only. From Wikipedia:

The first practical controllable-pitch propeller for aircraft was
  introduced in 1932. French firm Ratier pioneered variable-pitch
  propellers of various designs from 1928 onwards, relying on a special
  ball bearing helicoïdal ramp at the root of the blades for easy
  operation.

This means all speed records were flown with fixed-pitch propellers, which needed to be optimised to the top design speed to coax as much thrust from the engines as possible. In turn, this means that most of the propeller blade was stalled at low speed, so take-off thrust was low. To achieve high speed, a high wing loading was needed and, in combination with low thrust, the take-off length of such an aircraft was beyond of the length of airfields back then.
On water, however, the maximum field length is close to infinity. Yes, drag with floats is especially high when transitioning from buoyancy to dynamic lift (the "hump"), but that can be overcome when that hump occurs at high enough speed so the propeller delivers enough thrust.

This makes a lot of sense. Wouldn't it have been conceivable enough though to use some kind of assisted-start system (catapult? tossing line? rocket boosters?) to get a land-based plane quickly to the speed needed to get the propeller out of stall? Or else, to take off in a salt flat or something?

@leftaroundabout: There aren't that many salt flats in Europe where the rules were made with respect to record flights back then. I do not know for sure but would suspect that catapults also would had violated the rules.

Very interesting reasoning. I understand the issues of drag while the pontoons are still in the water, but are they not also a significant source of drag at altitude? In light of all the efforts gone into streamlining the fuselage, one would imagine that the presence of pontoons would provide a very relevant contribution to the total drag on plane?

@Monolo pictures f the S6B suggest yes, don't they

@Monolo: Yes, the floats and especially the sharp edges on their bottom and  the steps did contribute a lot to drag. That is why the Italian Piaggio P.7 tried to replace them with hydrofoils.

@Monolo: The top side of the floats doubled as area for radiators. They were not completely useless once the aircraft was in the air.

This may only be part of the story, but the Schneider Trophy was the most prestigious air trophy of that era, and it was initially reserved for seaplanes only. In I Could Never Be So Lucky Again, the famous aviator Jimmy Doolittle (writing about events in 1925) said:

The Schneider Cup race, to be held at Baltimore two weeks after the
  Pulitzer [air race], was reserved for seaplanes only. The race had first been run
  in 1913 and was sponsored by Jacques P. Schneider, pilot son of a
  wealthy French industrialist. It was considered the most important
  international air race at the time and received much press coverage in
  the flying nations of the world.

So at that time, the way to bring the most attention to your aircraft - or your country's aviation capabilities - was to enter the most famous air race, which happened to be restricted to seaplanes only:

The word came down from Washington that we could attempt to set or
  break records and make cross-country flights that were considered
  newsworthy and would project a favorable image of the Army Air
  Service.

I'm sure there were other factors involved, but publicity and prestige seem to be two big ones.

According to the book "Race with the Wind: How Air Racing Advanced Aviation" by Birch Matthews, "Schneider thought that seaplanes had a great future since so much of the earth is covered by water, which could be used for takeoff by large, heavy aircraft without the need to build runways." (Wikipedia)
I've also read somewhere (but cannot remember the source, sorry) that a further argument was that most settlements at least in the western world are constructed near sources of water, so seaplanes made sense as they could use existing infrastructure (docks) rather than building new airstrips.

This is the real reason.

@bjelleklang Re the "existing infrastructure", that's why seaplanes were the most popular large plane before WWII, and why wheeled planes were the most popular after WWII.  Almost all significant airports in Europe and much of the rest of the world were originally constructed as military airbases, and that meant the infrastructure for civilian aviation was already in place at the end of the war.  It's reasonable to assume that airports on land would have been constructed anyway without the war, but that was a major reason for the step-change.

No source on this, but my gut feeling is that it's a funding issue. The Air Force wasn't an independent branch and the navy was quicker to invest in aviation and didn't share with the army, and definitely had a larger budget then private citizens. The army definitely delayed emphasizing their Air Force (based on Rindlesbacher/rickenbacker's experiences). So again, no source other than books like The Aviators, but I would say funding in the Navy was higher and aircraft carriers weren't yet perceived by the navy as a whole as the dominant force they became in the early years of world war 2 (hence the pontoons).

You might want perhaps elaborate on Eddie Rickenbacher (if he's who “rickenbacher's experiences” refer to). But, more importantly, please correct me if I am wrong: I feel that this answer addresses mainly the American situation, while the years of the apogee of the seaplanes in the 1930s involve also, and perhaps mainly, British and Italian planes, designers and pilots. So perhaps the answer should be widened a bit, as well as sourced.

Excellent point. Best to delete my answer? I wanted to leave it as a comment, but apparently I can't yet comment on aviation questions, only answer...

At no point in this answer do you deign to mention what country you're talking about.

authors: Vladimir F Героям слава


text:
Would a ramjet be able to fly at 1% of sea level atmosphere and pure oxygen?
1% (0.01 atm) of the sea level pressure is a bit less pressure than at the SR-71 Blackbird's highest altitudes. If all that atmosphere was 100% oxygen (instead of 21%), would a (sc)ramjet still be able to fly in it? If not, would it be able to fly in it if Earth's gravity was lower?

Downvoter please explain why -1 so the question may be improved.

@mins Alright I rephrased it so that it doesn't sound like more appropriate to space.SE.

"This question does not show any research effort [...] or not useful". OP has not demonstrated they made any attempt at answering this question, and I don't see how this question is useful to the broader community.

@Sanchises You're right, I made an attempt to ask the question instead. What should I have done to answer it? The question and its answer should be useful to the OP in the first place.

Do you mean, would the engine be able to operate, or would the aircraft be able to fly? VTC as unclear.

@RalphJ Ability to operate but I'd be curious on the ability leveled flight too. The duplicate question answers it quite well anyway.


text:
What do you call a fall that is due to too little engine power, not due to cessation of aerodynamic lift?
If you raise your plane's pitch but the engines can no longer counter the gravity and you begin to fall (because the engines are too weak, not because of too little lift), what do you call that? This could also happen on planets without atmospheres (to rocket-propelled craft). Is it called a "stall" too, a different kind of stall? Or doesn't it have any special designation?

Comments are not for extended discussion; this conversation has been moved to chat.

"Begin to fall"  describes acceleration, which affects velocity.  This is why "Captain Kirk" and his three friends experienced weightlessness (while they were still climbing) as soon as their thrust (upward force) went to 0.  If they were still in (significant) atmosphere,  gravity + drag would have been greater than 1 G (downwards), they wouldn't float, they would hit the cieling of the capsule.

A "stall" may be terminology going back to the very origins of flight, when early aviators did not fully understand why there was a massive increase in drag (and accompanying loss of airspeed) when AoA exceeded a certain limit.
The AoA definition lives on today, but it is interesting to consider airspeed loss due to insufficient thrust or excessive vertical pitch.
The result of airspeed loss under more benign conditions is simply the mechanism of static stability.  A properly designed aircraft will pitch down before it stalls to maintain airspeed.  This is a function of CG placement and tail volume.  As the aircraft loses airspeed, it's flight path should begin to curve downwards, gaining airspeed.
In more extreme cases, such as a failed loop, the aircraft can lose all airspeed and even fall backwards.  But because it will then be highly directionally unstable, manuevers such as the "hammerhead" will allow safe recovery.
The inability to climb on all worlds is "insufficient thrust", and is also defined as "service ceiling".

Word stall was not invented by early aviators, but rather adopted by them. It's original meaning from the 1400's was "to come to a stand": https://www.etymonline.com/word/stall lists firs lists first use in aviation at 1910, and to describe the stopping of an engine at 1904, unfortunately the site does not list sources for these pieces of information. Stall quite neatly describes the seizure of flight in the same manner as an engine would stall.

An excellent example of applications.

In this case it's simply called: fall.
In aviation, stall is associated with lift-driven machines, and it seems from your question that you are not talking about this type of machinery.
You can find the definition of stall in this pdf document: Glossary of Terms from
Flight Research: Problems Encountered and What They Should Teach Us, by Milton
Thompson with J. D. Hunley, NASA:

Stall - loss of lift due to an acute wing angle, often caused by insufficient speed
stall - A condition in which an aircraft or airfoil experiences an interruption of airflow
resulting in loss of lift and a tendency to drop.

Notice that both definitions involve lift loss.
You can find many scientific papers that use the term "fall".

But a fall due to a too weak engine. So it's called insufficient thrust.

@Giovanni: What happens when X has insufficient thrust? it falls. If there is insufficient lift that's called stall.

@Giovanni: I have updated the answer.

You can fall due to many reasons.

@Giovanni: You can stall due to many reason too.

No, you stall because your aoa is at critical angle.

more on the subject of stall AoA: https://aviation.stackexchange.com/a/3620/42636

@Giovanni: Stalling  due to a critical angle of attack is fuzzy because that AoA is not fixed and depends on may parameters such as Re number. In dynamic stall encountered by Helicopters, Wind turbines it become very compilcated, because stall depends on many parameters (Reduced frequency, shape of the airfoil, Re number, Mach number, type of motion, etc.).

It's not a free fall because the engine is still running, the term "descent" would be better perhaps. Last sentence in my question description: "Or doesn't it have any special designation?" Your answer then is "No, it doesn't"?

A stall is a stall.  When the angle of attack exceeds critical, turbulent drag rises sharply and lift no longer rises with it.  It doesn't necessarily imply that you fall; a kite (like a common child's toy) normally flies in a stalled condition.
A stall can't happen without atmosphere, because the term implies that aerodynamic lift and drag are changing and that angle of attack exists (which it doesn't without air to flow over the wing).

Hence my question, I'm asking on a non-aerodynamic stall. One could call it a gravitational stall. A rocket tries to fly up but its engines aren't powerful enough, what would one call it.

There is no such thing as a "non-aerodynamic" stall.  The little tailstand in Alaska a few weeks ago (due to only four out of five engines igniting) wasn't a stall, it was insufficient thrust to weight ratio.  Words have meanings, and in aeronautics, "stall" means you exceeded critical angle of attack and lift stopped increasing in proportion to drag.

Your answer then is that one doesn't call the thing I described a "stall". Here comes the question: What does one call it? Yes I mean insufficient thrust to weight ratio and I wonder if it has a certain name when the rocketplane is falling due to it.

That would be a question for Space Exploration SE.  I'm used to hearing it called a "bad day."

How about "fall"

@Giovanni Can be referred to as engine stall in an airplane but in case of rocket maybe low thrust. There is nothing like a stall in a rocket at least I have never heard about it.

@Giovanni You just call it "insufficient thrust". That's it.

@Jpe61 But the engines are on which doesn't make it a free fall.

I didn't suggest "free"

"If you raise your plane's pitch but the engines can no longer counter the gravity and you begin to fall (because the engines are too weak, not because of too little lift), what do you call that?"
At this moment you've described, you DO have "too little lift"- otherwise you wouldn't be falling.
If the wing is stalled (not flying) due to insufficient relative wind going over the airfoil (for whatever reason) you are in a stalled condition.
An airplane at altitude doesn't need an engine or thrust to fly, it only needs a pilot to keep the nose (pitch attitude) low enough that the wing's critical angle of attack is not being exceeded, thereby ensuring that the airflow remains attached to the upper surface of the wing creating lift.
Same as a glider.
A falling airplane is considered to be in the "stalled" condition, regardless of how or why the pilot mismanaged the airspeed or pitch attitude to put the aircraft in that condition.
If you put the nose down it will become un-stalled once you reach the wing's minimum flying speed (exceed stall speed).

If you raise your plane's pitch but the engines can no longer counter
the gravity and you begin to fall (because the engines are too weak,
not because of too little lift), what do you call that

Well, for starters, a steady-state descent happens when thrust is less than drag, but equal to drag minus the component of the weight vector that is acting parallel to the flight path. A steady-state climb happens when thrust is greater than drag, but equal to drag plus the component of the weight vector that acts parallel to the flight path.  (Also, to maintain the steady-state condition, the lift vector must be appropriate as well-- the appropriate value is equal to weight * cosine (climb angle or glide angle).  Otherwise, the flight path will curve up or down-- which will typically change the airspeed as well, by upsetting the balance between thrust, drag, and the component of the weight vector acting parallel to the flight path.
Is a steady-state descent a type of falling?  Is a steady-state descent which happens to be straight downward, with the wings at the zero-lift angle-of-attack and the airspeed at terminal velocity, a type of falling?
Or does "falling" only refer to a downward curve of the flight path?  Or is a better definition any situation where the aircraft's vertical speed is growing more negative or less positive?  Or is it simply any situation where we find ourselves to be travelling rapidly towards mother earth?
Your question doesn't appear to be referencing a steady-state condition. Maybe what you mean by "fall" is the situation where you gain some extra airspeed and then you arc the flight path up into a steep climb and try to keep the airplane climbing along that line, but you don't have enough thrust to maintain airspeed-- since to keep the airspeed constant, the thrust must oppose not only the drag force, but also the component of gravity which is acting parallel to the flight path.  Otherwise, something has to "give".  Now the situation could unfold in several different ways, depending on the pilot's control inputs.

The pilot keeps pulling the control stick back to try to maintain the linear climb, and eventually the wing is placed at the stall angle-of-attack and the wing stalls.

The pilot keeps pulling the control stick back to try to maintain the linear climb, but as he approaches the max L/D ratio he realizes this isn't going to work so he refrains from pulling the stick back any further, so the wing's angle-of-attack remains constant after this point.  As the airspeed (and therefore the lift vector) decreases further, the flight path eventually starts arcing downward.  What happens next could range from a gentle transition back to steady-state flight along a more sustainable flight path, be it climbing, descending, or level, or a rather abrupt pitch-over from a steep climb to a steep dive, depending on how extreme the initial climb was.  There certainly could be a "falling" sensation involved, even if the wing never reaches the stall angle-of-attack (and in some cases even if the aircraft never actually enters a descent).  Think of a rocket being blasted straight up on a vertical trajectory, without enough energy to escape the atmosphere-- what happens at the top of that trajectory might be quite exciting, even if it has nothing to do with a wing reaching the stall angle-of-attack.

At some point after establishing the linear climbing trajectory, the pilot starts moving the control the stick forward as needed to hold airspeed constant, rather than allowing it to further decrease.  The flight path will arc downward into a more sustainable trajectory, be it climbing, descending, or level.  The wing will certainly not reach the stall angle-of-attack, but the arcing transition could in a sense be described as "falling".  An interesting exercise would be to precisely describe how the wings' lift vector changes throughout this maneuver-- it cannot stay constant, since the angle-of-attack is being decreased but the airspeed is not increasing. The G-meter mounted on the panel would show a decrease during the push-over maneuver.  Depending on the airspeed at the time the pilot begins moving the control stick forward, it may or may not be possible to get the plane stabilized on a steady-state trajectory without ever having the airspeed increase, but in all cases the airspeed could be held constant for a while as the aircraft floats "over the top" of the change in trajectory.

We could undoubtedly conceive of several more permutations, again depending on exactly what control inputs the pilot makes.

So, the answer to your question is "it depends on what the pilot does".  But examples like rockets, bullets, thrown rocks, etc show that an object certainly can "fall" without having wings that are meeting the air at or above the stall angle-of-attack!
On a more practical (and less verbose) level, consider the case where a pilot "unloads" the wing to the zero-lift angle-of-attack to avoid a stall.  This maneuver can be carried out regardless of whether the flight path was climbing, descending, or level at the time the aircraft approached the stall, and regardless of whether the aircraft's airspeed was decreasing or increasing as the aircraft approached the stall.  (Consider steeply banked, tightly turning cases as well.)   What happens next?  A "fall".
Note however that in all cases, even including wingless objects, the only way that the flight path can arc downward into a different trajectory (be it climbing, descending, or level, and regardless of how abrupt the downward arc is) is if the actual lift vector has become smaller than the lift vector that is required to sustain that trajectory.  The required lift vector is equal to weight * cosine (climb angle or glide angle) -- as noted in this related ASE answer.  (We're now assuming wings-level (non-turning, non-banked flight for simplicity.) In a very steep climb (or a very steep dive), this value will be much less than weight, but if the wings aren't making that much lift (or if there are no wings!), the flight path will still arc (curve) downward.  If the wings are making that much lift, the flight path can't arc (curve) downward.  So your thesis about "begin to fall (because the engines are too weak, not because of too little lift)" doesn't make sense.  Regardless of how weak or strong the engines are-- regardless of whether the thrust-to-weight ratio is 10-to-1, 1-to-1, 1/10-to-1, or 0 (as in a glider), the only way the flight path can arc downward ("fall") is if the lift vector is smaller than the lift vector that is required to sustain the trajectory that exists at that instant, and that required lift vector is equal to weight * cosine (climb angle or glide angle.)
The bottom line is that extreme climbing maneuvers conducted with weak engines tend to cause place an aircraft in a position where the lift vector is less than weight * cosine (climb angle), and the flight path therefore curves downwards.  This happens because lift depends on airspeed as well as angle-of-attack, and if you are steeply climbing with a weak engine, you can't maintain airspeed.  So you lose lift, regardless of whether you place the wing anywhere near the stall angle-of-attack or not.
Standard caveats apply-- for simply, we're assuming that the thrust line is parallel to the instantaneous flight path, and except where stated otherwise, all actual formulae are intended for the simple case of non-turning (i.e. non-banked) flight.

I probably ought to just remove all the content except the last two major (bolded) paragraphs-- that's the real answer.

Next ASE question-- "What is falling?"  -- I'd argue that the word has some ambiguity, and that anyone using it in a question really ought to include a definition.  I suppose the most logical definition is "moving rapidly earthward", and therefore whether or not an aircraft ends up "falling" (at least temporarily) in any of the scenarios 1 through 3 in this answer, or other related scenarios, depends largely on how steep the initial climb was.

“I probably ought to just remove all the content…”. I agree.  @Sanchises comment is all this question needed.


text:
How to most likely get a flight that will go above 40,000 ft?
I wonder if there are passenger flights that would reach cruise altitudes higher than 40,000 ft (i.e. reach 41,000 ft for instance) more likely than others. It also depends on the plane of course, but assuming one gets a plane that can fly higher than 40k ft, e.g. the B777 or the B787 Dreamliner, are there other factors contributing to the likelihood of flying at higher altitudes?
The highest I think I've ever been is 38,000 ft, this was on a flight from Djerba to Central Europe in a B737 I think, but I'm unsure whether I didn't reach 39,000 ft on a transatlantic flight once in a B747 (it's too long ago to remember well). Then I've flown on many flights where there weren't those onboard screens where you can read altitude, speed and temperature but the captain announces the cruise altitude sometimes. However, I'm quite sure I've never flown above 40k ft (12.2 km). Are north-south flights more likely to fly at higher altitudes than west-east flights and/or are there other factors that may contribute to flying higher?
Edit: Following the revelations here I'm now sure that the flight from Djerba reached FL370, not 380 (it went slightly eastwards), and the transatlantic flight which was westwards to America probably reached FL380. So 38,000 ft MSL is really the highest I've ever been unless I don't remember correctly or haven't looked on a screen while a plane was higher or have been higher in some flight without those onboard screens and without the captain mentioning the cruise altitude.

Related: Why are many jet aircraft designed to cruise around FL350-370? and Why then don't aircraft fly even higher, for even greater efficiency? and

Welcome to the site. @Giovanni, this is not a forum where you ask people to share experiences, it's a question and answer site. I don't see a clear question here, if you have a specific thing you want to know please edit, if you are trying to start a general discussion this isn't the place for it.

@GdD Your're obviously referring to the last sentence only. My question is far from being a forum question, it asks which (and if any) flights would be more likely to go above 40k ft, which factors contribute to that.

@GdD I simply removed the offending sentence. I agree that it's not appropriate for this board, however the rest of the question is (potentially) reasonable.

I guess the above three comments can be deleted now.

Within the US, the recipe would be something along these lines:

Eastbound

Since FL 410 is an eastbound altitude

Moderately short, 1-2 hours
- not so short that climbing high isn't worth it, nor so long that the aircraft takes off so heavy with fuel that 410 can only happen as a step-climb
An aircraft that can reach 410

For example, a 737 NG or Max or a 777
Not an A-320 nor most regional jets
Ideally, for 737s, a 737-700, since this is a lighter airframe than the longer -800 or -900

Light load of passengers

If total aircraft weight is high, the highest altitudes are unachievable, so a 737-700 that is half full can go higher sooner than a full one
An early morning or late night flight may be the best place to start looking for a light load; a mid-day flight in a popular market is more likely to be full (or close to it)

Smooth air at 410

Examples of things to avoid here would be crossing the Rocky Mountains, especially the front range, especially with high winds (i.e. winter months), or crossing a major frontal boundary
Mountain wave becomes a problem when the airspeed variation it causes outstrips the performance of the aircraft to compensate, and performance is reduced at high altitudes.

Unless you're looking specifically to book a flight for the purpose of getting to 410, a lot of these factors are pretty "luck of the draw," but these things tend to be pretty common among the flights that do go that high.

Well, you can combine a touristic visit destination with the recommendations above. Thank you. I hope the same are true for Europe. But what do you mean FL410 is an "eastbound altitude"? Why does it matter whether you fly to the west or east? Can I expect a flight from the British Isles to Central Europe would more likely reach 41k ft than opposite direction?

@Giovanni The same is true in Europe (regardless of RVSM or not since RVSM ends at FL410). See the Wikipedia article on flight levels.

@Bianfable Alright, so the opposite direction could allow for an even higher FL such as 430 if I'm correct.

My bad, I forgot (IANAP) that above FL400, it's 2000 foot separation, not 1000 feet. Deleting so as to not spread false info.

@Giovanni  If you're in an aircraft capable of flying at FL 430, then yes. However, I don't think you'll find any airliner that can (legally -- per manufacturer certifications) go higher than FL 410 these days.  (The Concorde certainly could, but for a flight in that, you'll first need a trip in a time machine.)

@RalphJ I hope I'd get an eastward-flight at FL410. However, many airliners have a service ceiling of 43,100 ft and somewhere on Quora an ATC said the highest flight they've seen was cruising at 43,000 ft. But it's more likely to cruise at 41,000 ft probably, also because more planes are capable of flying there. One more question: How high are mountains that would be a concern to the smooth air? I guess you don't have to worry about mountains lower than ~6,000 ft, have you?